Problem: Evaluate $\dfrac{d}{dx}\sqrt{3-7x-x^2}$ at $x=-1$. Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac94$ (Choice B) B $3$ (Choice C) C $-\dfrac56$ (Choice D) D $\dfrac16$
Explanation: Let's start by finding the expression for $\dfrac{d}{dx}\sqrt{3-7x-x^2}$. Then, we can evaluate it at $x=-1$. $\sqrt{3-7x-x^2}$ is a radical expression, but its argument isn't simply $x$. Therefore, it defines a composite radical function. In other words, suppose $u(x)=3-7x-x^2$, then $\sqrt{3-7x-x^2}=\sqrt{u(x)}$. $\dfrac{d}{dx}\sqrt{3-7x-x^2}$ can be found using the following identity: $\dfrac{d}{dx}\sqrt{u(x)}=\dfrac{1}{2}[u(x)]^{^{-\scriptsize\dfrac{1}{2}}}u'(x)$ [Why is this identity true?] Let's differentiate! $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\sqrt{3-7x-x^2} \\\\ &=\dfrac{d}{dx}\sqrt{u(x)}&&\gray{\text{Let }u(x)=3-7x-x^2} \\\\ &=\dfrac{1}{2}[u(x)]^{^{-\scriptsize\dfrac{1}{2}}}u'(x) \\\\ &=\dfrac{1}{2}[3-7x-x^2]^{^{-\scriptsize\dfrac{1}{2}}}(-7-2x)&&\gray{\text{Substitute }u(x)\text{ back}} \end{aligned}$ Now let's evaluate $\dfrac{d}{dx}\sqrt{3-7x-x^2}$ at $x= -1$. $\begin{aligned} &\phantom{=}\dfrac{1}{2}\Bigl(3-7( {-1})-( {-1})^2\Bigr)^{^{-\scriptsize\dfrac{1}{2}}}\cdot\Bigl(-7-2({-1})\Bigr) \\\\ &=\dfrac{1}{2}\cdot 9^{^{-\scriptsize\dfrac{1}{2}}}\cdot(-5) \\\\ &=\dfrac{1}{2}\cdot \dfrac13\cdot (-5) \\\\ &=-\dfrac56 \end{aligned}$ In conclusion, the value of $\dfrac{d}{dx}\sqrt{3-7x-x^2}$ at $x=-1$ is $-\dfrac56$.